[tex]Problema~este~simpla:~Pentru~a~demonstra~ca~n^2+5n+6~nu \\ \\ este~patrat~perfect,~trebuie~sa~il~incadram~intre~doua~patrate~ \\ \\ perfecte~consecutive. \\ \\ Pentru~asta~ne~folosim~de~coeficientul~lui~n,~adica~5:~4\ \textless \ 5\ \textless \ 6. \\ \\ 4=2 \cdot 2~si~6=2 \cdot 3...Asta~imi~sugereaza~sa~verific~daca~numarul~se \\ \\ incadreaza~intre~(n+2)^2~si~(n+3)^2.[/tex]
[tex]Rezolvarea~propriu-zisa: \\ \\ Intr-adevar:~(n+2)^2=n^2+4n+5\ \textless \ n^2+5n+6. \\ \\ De~asemenea~(n+3)^2=n^2+6n+9\ \textgreater \ n^2+5n+6. \\ \\ Din~aceste~doua~relatii~rezulta:~(n+2)^2\ \textless \ n^2+5n+6\ \textless \ (n+3)^2, \\ \\ si~mai~apoi:~n+2\ \textless \ \sqrt{n^2+5n+6}\ \textless \ n+3. \\ \\ Si~desigur,~daca~ \sqrt{p} \in Q-Z~(unde~p \in N)~atunci~ \sqrt{p} \in R-Q.~ \\ \\ (Adica~ \sqrt{p}~cu~p \in N~este~fie~natural,~fie~irational) [/tex]
